Answer
Local minimum at $f(0,0)=0$ and saddle points are at $(\pm 1,0)$
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum, local maximum, or a saddle point.
For $(x,y)=(0,0)$
$D(0,0)=4 \gt 0$ and $f_{xx}(0,0)=2 \gt 0$
For $(x,y)=(-1,0)$
$D(-1,0)=-16e^{-2} \lt 0$
For $(x,y)=(1,0)$
$D(-1,0)=-16e^{-2} \lt 0$
Hence,
Local minimum at $f(0,0)=0$ and saddle points are at $(\pm 1,0)$